Odds Generation — soccer · goal · regular

Enter a period's Handicap and Total in Malay and its 1X2 in decimal — margins included. The engine:

1. strips the margins (Power for the two-way pairs, Shin for the 1X2) to fair probabilities;
2. back-solves the goal rates (λₕ, λₐ) by nested bisection — Total fixes how many goals, Handicap fixes who scores them;
3. refines the score grid in stages — Poisson baseline, Dixon-Coles ρ, optional φ draw lift;
4. prices every market off the last stage.

Stage 1 — (λₕ, λₐ) outer-product into the joint score grid

Flip Poisson / Dixon-Coles / φ on the grid to see what ρ and φ move.

Each period — full match, 1st half, 2nd half — is its own inversion. Timing and cross-half markets (HT/FT, First Goal, …) are out of scope; see the Markets reference.

Primary Quotes

Handicap (Malay)

LineHomeAway

Total (Malay)

LineOverUnder

Dixon-Coles

ρ

1X2 quote (Decimal) optional · φ draw fit

Home1DrawXAway2

↑/↓ steps a field, Shift steps bigger; on a Malay pair the partner moves the opposite way. Handicap/Total ladders use your line ±0.25; team-total, 3-way and which-team lines are illustrative — probabilities are exact given λ.

Valid inputs: Malay in [−1, +1] and never 0, 1X2 decimals above 1, and each book must carry margin (implied probabilities summing over 1).

λₕ 1.7339λₐ 1.0174λ_total 2.7512φ 0.2042round-trip Δp 1.7e-9 / 2.1e-8 / 2.0e-10draw 0.2706 → 0.2995grid N=5 · Poisson → Dixon-Coles → 1X2 φ

Score grid — P(home = h, away = a) ×100

Poisson

λₕ 1.6378λₐ 1.0670draw 0.2459

	A=0	A=1	A=2	A=3	A=4	A=5
H=0	6.69	7.14	3.81	1.35	0.36	0.08
H=1	10.95	11.69	6.24	2.22	0.59	0.13
H=2	8.97	9.57	5.11	1.82	0.48	0.10
H=3	4.90	5.23	2.79	0.99	0.26	0.06
H=4	2.01	2.14	1.14	0.41	0.11	0.02
H=5	0.66	0.70	0.37	0.13	0.04	0.01

home Σ 0.5000 draw Σ 0.2459 away Σ 0.2465

Dixon-Coles

λₕ 1.6469λₐ 1.0427draw 0.2706

	A=0	A=1	A=2	A=3	A=4	A=5
H=0	8.02	5.96	3.72	1.29	0.34	0.07
H=1	10.09	12.93	6.13	2.13	0.56	0.12
H=2	9.28	9.68	5.04	1.75	0.46	0.10
H=3	5.09	5.31	2.77	0.96	0.25	0.05
H=4	2.10	2.19	1.14	0.40	0.10	0.02
H=5	0.69	0.72	0.38	0.13	0.03	0.01

home Σ 0.5000 draw Σ 0.2706 away Σ 0.2294

1X2 φ

λₕ 1.7339λₐ 1.0174φ 0.2042draw 0.2995

	A=0	A=1	A=2	A=3	A=4	A=5
H=0	8.67	5.15	3.17	1.07	0.27	0.06
H=1	9.53	14.30	5.49	1.86	0.47	0.10
H=2	9.20	9.36	5.73	1.61	0.41	0.08
H=3	5.32	5.41	2.75	1.12	0.24	0.05
H=4	2.30	2.34	1.19	0.40	0.12	0.02
H=5	0.80	0.81	0.41	0.14	0.04	0.01

home Σ 0.5000 draw Σ 0.2995 away Σ 0.2005

shade ∝ probability · each stage re-solves (λₕ, λₐ) against the same fair targets

MarketHandicapOver/UnderHome Over/UnderAway Over/Under1X2Correct ScoreDraw No BetOdd/EvenHome Odd/EvenAway Odd/EvenMyanmar HandicapMyanmar Over/UnderDouble ChanceClean SheetTo Win to NilBoth/One/Neither Teams to ScoreWinning Margin3-Way HandicapExact TotalRange TotalWhich Team Will Win By X+21 in scope

Multi-way (Shin)

1-01019

2-01130

2-11017

3-01883

3-11850

3-23457

4-041270

4-140172

4-276194

4-3196301

0-0

11

1-1

6.8

2-2

17

3-3

80

4-4

482

AOS

37

How these numbers are computed

1 · Strip the bookmaker margins — Power (two-way) · Shin (three-way)

A two-way Malay price converts to decimal ($d = 1 + m$ if $m > 0$, else $d = 1 - 1/m$), which implies a probability $q = 1/d$; the two sides sum to an overround $Q > 1$. The Power method finds the exponent $x$ that deflates them to a fair pair $p_i = q_i^{1/x}$, and the stripped (margin-free) decimal odds are the reciprocals $1/p_i$. The three-way 1X2 needs a different deflation — Shin (the formula is in step 5) — but lands in the same table shape. The fair numbers are the constraints the solver must reproduce.

Handicap -0.50 → P(home covers -0.50)

	Home	Away
Malay quote	+0.96	+0.96
Decimal $d$	1.9600	1.9600
Implied $q = 1/d$	0.5102	0.5102
Fair $p = q^{1/x}$	0.5000	0.5000
Fair decimal $1/p$	2.0000	2.0000

overround Q = 1.0204 → solve $q_1^{1/x} + q_2^{1/x} = 1$ → x = 0.9709 → fair P(home covers) = 0.5000

Total 2.50 → P(over 2.50)

	Over	Under
Malay quote	+0.95	+0.95
Decimal $d$	1.9500	1.9500
Implied $q = 1/d$	0.5128	0.5128
Fair $p = q^{1/x}$	0.5000	0.5000
Fair decimal $1/p$	2.0000	2.0000

overround Q = 1.0256 → solve $q_1^{1/x} + q_2^{1/x} = 1$ → x = 0.9635 → fair P(over) = 0.5000

1X2 (Decimal) → P(home win), P(draw), P(away win) — Shin

	Home	Draw	Away
Decimal quote $d$	1.8900	3.1300	4.7000
Implied $q = 1/d$	0.5291	0.3195	0.2128
Fair $p$ (Shin)	0.5060	0.2995	0.1945
Fair decimal $1/p$	1.9762	3.3389	5.1420

overround Q = 1.0614 → bisect Shin’s insider fraction z until $\textstyle\sum p_i = 1$ (step 5) → z = 0.0308 → fair draw = 0.2995, the φ target

Try the strips standalone — Margin — two-way (Power) · Margin — multi-way (Shin): the same math on any quotes, with the bisection narrated.

2 · Size — recover λ_total from the Total

Step 1 left three fair targets, and the model has matching knobs — the scoring rates $\lambda_h, \lambda_a$ plus the diagonal lift φ (step 5). Model each side’s goals as Poisson — $P(k;\lambda) = e^{-\lambda}\lambda^{k}/k!$ — and independent. Independent Poissons add — $X + Y \sim \mathrm{Poisson}(\lambda_h + \lambda_a)$ — so the match total depends only on $\lambda_{\text{total}} = \lambda_h + \lambda_a$: the Total quote pins the match size on its own, before caring who scores (step 3’s job). Every line class settles by the same rule (the line rules below): take the win mass $W$ and push mass $R$ of that one total — point masses $P(\text{total} = t) = e^{-\lambda}\lambda^{t}/t!$ — and the fair over-probability is $W/(1-R)$. That makes size a one-dimensional root-find, laid out below in the order a solver runs it — target, search, then the evaluation every iteration repeats:

target (step 1): $$p^{\ast}_{\text{over}} \;=\; \text{fair } P(\text{over } 2.50) \;=\; 0.50000000$$

search: $$\text{bisect } \lambda_{\text{total}} \in [0.02,\, 8] \;\text{ until }\; W/(1-R) = p^{\ast}_{\text{over}} \;\Rightarrow\; \lambda_{\text{total}} = 2.704832$$

Cross-check in the Bisection Calculator: paste Ptail(2.50, x) = 0.50000000 (tolerance 1e-8; the target needs its full precision — $d\lambda/dP \approx 4$ here). Its search lands on $\lambda_{\text{total}} = 2.674060$ — close to, but not the $2.704832$ above: $\Delta = 3.1 \times 10^{-2}$. The calculator inverts the closed-form total $X + Y \sim \text{Poisson}(\lambda_{\text{total}})$, while this stage inverts the score grid truncated at N = 5 (no renormalization), which clips the upper tail. That gap is a model difference, not rounding — the 6-dp $\lambda_{\text{total}}$ above is what a faithful N = 5 re-implementation reproduces; the two agree only where the truncated tail is negligible.

Each trial λ is scored the same way — the settlement rule evaluated under Poisson(λ). These are the figures a re-implementation should reproduce at the converged $\lambda = \lambda_{\text{total}} = 2.704832$: first each component line’s Poisson masses, then the weighted $W$ and $R$, then the fair check against the target:

$$P(\text{total} > 2.50) \;=\; 1 - \textstyle\sum_{t=0}^{2} \tfrac{e^{-\lambda}\lambda^{t}}{t!} \;=\; 0.5000 \qquad P(\text{total} = 2.50) \;=\; 0 \;\; \text{(totals are integers)}$$

$$W = P(\text{total} > 2.50) = 0.5000$$

$$R = P(\text{total} = 2.50) = 0.0000$$

$$\text{fair } P(\text{over } 2.50) \;=\; \frac{W}{1-R} \;=\; \frac{0.5000}{1 - 0.0000} \;=\; 0.5000 \;=\; p^{\ast}_{\text{over}} \;\; \checkmark$$

$$\Longrightarrow\;\; \lambda_{\text{total}} = 2.704832$$

(Additivity is exact only for the plain independent-Poisson grid, so this λ is the Poisson stage’s; the τ and φ adjustments of the later stages perturb it, so each re-solves size and split jointly against the same targets — all the recovered λ’s land in the per-stage tables (steps 4–5).)

Line rules — over wins ⟺ $h + a > T$; an exact hit ($h + a = T$) is a push: the stake refunds, so a two-way fair probability renormalizes it away, $\text{fair} = W/(1-R)$ (win mass over non-push mass). A .5 line cannot push, so $R = 0$ and the formula degenerates to $\text{fair} = W$ — same rule, not a shortcut method; a quarter line (±0.25, ±0.75, …) puts half the stake on each neighbouring line, so $W$ and $R$ are the ½-weighted sums over the two components. The same settlement applies to the Handicap (step 3) and the priced ladders (step 6).

3 · Split — recover λₕ and λₐ from the Handicap

The Handicap decides who scores them — and settling it needs the full scoreline distribution, the score grid: independence makes each cell $(h, a)$ a plain product of the two Poisson masses,

$$J[h][a] \;=\; P(\text{home} = h;\, \lambda_h) \cdot P(\text{away} = a;\, \lambda_a) \qquad h, a = 0 \ldots 5$$

— exactly the grid the score-grid panel’s Poisson stage displays. Hold $\lambda_{\text{total}}$ fixed and slide the home share $\lambda_h$ (with $\lambda_a = \lambda_{\text{total}} - \lambda_h$). Home covers ⟺ $h + L > a$, an exact $h + L = a$ is a push — step 2’s one rule again, with the masses now read off the grid: $W = \textstyle\sum_{h + L > a} J[h][a]$ and $R = \textstyle\sum_{h + L = a} J[h][a]$ (½-weighted across a quarter line’s two components), fair cover probability $W/(1-R)$ — which rises with $\lambda_h$, so the same solver shape applies:

target (step 1): $$p^{\ast}_{\text{cover}} \;=\; \text{fair } P(\text{home covers } -0.50) \;=\; 0.50000000$$

search: $$\text{bisect } \lambda_h \in (0,\, \lambda_{\text{total}}) \;\text{ until }\; W/(1-R) = p^{\ast}_{\text{cover}} \;\Rightarrow\; \lambda_h = 1.637850,\;\; \lambda_a = \lambda_{\text{total}} - \lambda_h = 1.066982$$

Cross-check in the Bisection Calculator — and a closer look at what $P(\text{home covers } -0.50)$ actually sums. Step 2's Total was a sum of the two goal counts (one rate, one Ptail); the Handicap turns on their difference — the goal margin $X - Y$ (home goals minus away goals). A margin depends on both rates at once — its distribution is a Skellam, the difference of two Poissons — so no single Ptail in $\lambda_{\text{total}}$ can express it. The trick is to condition on the away score $a$: once $a$ is pinned, only the home goals still vary, and the cover collapses to one plain home tail ($\mathrm{Ptail}(k;\, \lambda) = P(X \ge k)$). Home covers $\iff h \ge a + 1$ (a half-line cannot push), so each away score contributes its Poisson mass times that tail:

$$P(\text{home covers } -0.50) \;=\; \sum_{a \ge 0} P(\text{away} = a;\, \lambda_a)\,\mathrm{Ptail}(a + 1;\, \lambda_h)$$

The goal margin $m = h - a$ — a Skellam, the difference of two Poissons. Shaded by how the $-0.50$ line settles: home covers ($m \ge +1$), away.

Now substitute the search variable $x = \lambda_h$ (with $\lambda_a = \lambda_{\text{total}} - x$, and $\lambda_{\text{total}} = 2.704832$ held fixed from step 2) so the whole cover is one function of $x$:

$$W \;=\; \sum_{a \ge 0} P(a;\, \lambda_{\text{total}} - x)\,\mathrm{Ptail}(a + 1;\, x)$$

$$\text{fair cover} \;=\; W \quad (\text{no push, so } R = 0)$$

Truncate the away sum at N = 5 and paste this equation — one line per away score $a$, interval $[0,\, \lambda_{\text{total}}]$, tolerance 1e-8:

  P(0, 2.704832-x)*Ptail(1, x)
+ P(1, 2.704832-x)*Ptail(2, x)
+ P(2, 2.704832-x)*Ptail(3, x)
+ P(3, 2.704832-x)*Ptail(4, x)
+ P(4, 2.704832-x)*Ptail(5, x)
+ P(5, 2.704832-x)*Ptail(6, x)
= 0.50000000

Its search lands on $\lambda_h = 1.623928$; at that $x$ the 6 terms evaluate and add straight up to the target ($\lambda_a = \lambda_{\text{total}} - \lambda_h = 1.080904$):

$$0.2724 + 0.1770 + 0.0442 + 0.0059 + 0.0005 + 0.0000 \;=\; 0.5000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

That $\lambda_h$ is close to, but not exactly, the $1.637850$ from the grid solve above — a gap of $\Delta = 1.4 \times 10^{-2}$. It is the same effect as the Total's: the calculator uses the full distribution (goals can run past 5 a side), while this page's grid stops at 5 and quietly drops the sliver of probability beyond it — the faded tail in the figure below. Dropping it leaves a hair less mass on the home side, so the grid nudges $\lambda_h$ up a touch to land on the same target; that nudge is $\Delta$. Where the dropped tail is negligible, the two agree.

One side's goal count (Poisson). The grid keeps 0–5 and drops the faded tail past the dashed cap — here just 0.67% of the mass. That sliver is the whole source of $\Delta$.

At the converged split the grid masses and the fair check are:

$$W = P(h - 0.50 > a) = 0.5000$$

$$R = P(h - 0.50 = a) = 0.0000$$

$$\text{fair } P(\text{home covers } -0.50) \;=\; \frac{W}{1-R} \;=\; \frac{0.5000}{1 - 0.0000} \;=\; 0.5000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

$$\Longrightarrow\;\; \lambda_h = 1.637850 \qquad \lambda_a = 1.066982$$

The recovered pair $(\lambda_h,\, \lambda_a)$ is the Poisson stage’s chips in the grid panel — and with it both quotes are reproduced exactly. That closes the fit, but not the model: the two quotes pinned just two numbers, size and split, while step 6 prices whole ladders off the grid’s shape — and that shape so far rests entirely on the independence assumption. The next steps correct the shape where independence is known to fail, then re-solve so the quotes still hold: Dixon-Coles’ low-score τ (step 4), then the 1X2 draw lift φ (step 5).

4 · Dixon-Coles — re-solve with the low-score tilt τ

Steps 2–3 hit both targets exactly — so what is left? Infinitely many grids satisfy those same two constraints; the independent product is merely the simplest of them, and it is empirically wrong in one known corner: real matches produce slightly more 0-0 and 1-1, and fewer 1-0/0-1, than independence predicts (the Dixon-Coles finding). The fix multiplies exactly those four cells by a tilt τ driven by a correlation knob ρ. Note ρ is not recovered from your quotes — steps 2–3 already spent both of them — it is a modelling input, the ρ field in the Dixon-Coles section above (league-calibrated in practice). Negative ρ lifts 0-0/1-1 and trims 1-0/0-1; positive ρ the reverse; τ = 1 leaves everything else alone:

$$J[h][a] \;\mapsto\; J[h][a] \cdot \tau(h,a) \qquad \tau = 1 \;\text{everywhere except the four low-score cells}$$

with ρ = -0.10 at the re-solved pair (λₕ, λₐ) = (1.6469, 1.0427) from below:

$$\tau(0,0) \;=\; 1 - \lambda_h \cdot \lambda_a \cdot \rho \;=\; 1 - 1.6469 \times 1.0427 \times (-0.10) \;=\; 1.1717$$

$$\tau(0,1) \;=\; 1 + \lambda_h \cdot \rho \;=\; 1 + 1.6469 \times (-0.10) \;=\; 0.8353$$

$$\tau(1,0) \;=\; 1 + \lambda_a \cdot \rho \;=\; 1 + 1.0427 \times (-0.10) \;=\; 0.8957$$

$$\tau(1,1) \;=\; 1 - \rho \;=\; 1 - (-0.10) \;=\; 1.1000$$

The tilted grid is renormalized to $\textstyle\sum J = 1$ (ρ is valid only while every τ stays positive). That reshaping breaks the step 2–3 fit — the same λ pair now prices differently, because $W$ and $R$ are sums over a grid whose mass just moved. Nothing else changes: the targets stay step 1’s, the settlement rule stays $W/(1-R)$, and the same two nested bisections simply run again with the τ-tilted grid underneath. In the same solver order as steps 2–3:

targets (step 1, unchanged): $$p^{\ast}_{\text{over}} = 0.5000 \qquad p^{\ast}_{\text{cover}} = 0.5000$$

old pair on the τ-grid: $$(\lambda_h,\, \lambda_a) = (1.6378,\, 1.0670): \quad \text{fair } P(\text{over}) = 0.5038 \qquad \text{fair } P(\text{cover}) = 0.4920 \qquad \text{both off target}$$

search: $$\text{bisect } \lambda_{\text{total}} \in [0.02,\, 8] \text{ and } \lambda_h \in (0,\, \lambda_{\text{total}}) \text{ exactly as in steps 2–3, } W \text{ and } R \text{ now summed off the } \tau\text{-grid}$$

check (new pair, τ-grid): $$\text{at } (\lambda_h,\, \lambda_a) = (1.6469,\, 1.0427)\!: \quad \text{fair } P(\text{over } 2.50) \;=\; \frac{W_{\tau}}{1-R_{\tau}} \;=\; \frac{0.5000}{1 - 0.0000} \;=\; 0.5000 \;=\; p^{\ast}_{\text{over}} \;\; \checkmark$$

$$\text{fair } P(\text{home covers } -0.50) \;=\; \frac{W_{\tau}}{1-R_{\tau}} \;=\; \frac{0.5000}{1 - 0.0000} \;=\; 0.5000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

$$\Longrightarrow\;\; (\lambda_h,\, \lambda_a): \; (1.6378,\, 1.0670) \;\xrightarrow{\;\tau,\;\rho \,=\, -0.10\;}\; (1.6469,\, 1.0427) \qquad \lambda_{\text{total}}: \; 2.7048 \to 2.6896$$

Step 3 ended on $\text{fair } P(\text{home covers } -0.50) = 0.5000$ with $\lambda_h = 1.6378$; step 4 ends on the same 0.5000 with $\lambda_h = 1.6469$. That is by construction, not coincidence: 0.5000 is the target — step 1’s $p^{\ast}_{\text{cover}}$ — and a bisection only stops when it lands there, so every ✓ line prints the same number no matter which grid sits underneath. The fair value is a function of two inputs, the grid and the λ pair; the λ’s differ precisely because the grid differs. Evaluate both pairs on both grids and the compensation becomes visible:

fair P(home covers -0.50)	product grid (step 3)	τ-grid, ρ = -0.10 (step 4)
step-3 pair (1.6378, 1.0670)	0.5000 ✓	0.4920 ✗
step-4 pair (1.6469, 1.0427)	0.5078 ✗	0.5000 ✓

Read it by row — same pair, swap the grid — and the tilt shows: at the step-3 pair, τ alone drags the cover value 0.5000 → 0.4920. Read it by column — same grid, swap the pair — and the repair shows: on the τ-grid, moving to the step-4 pair brings it back to 0.5000. Each grid has exactly one ✓ cell; tilting the grid moved the solution point, and finding its new location is all the step-4 bisection does. ρ itself never enters the settlement rule — it acts one layer down, inside the cells that $W$ and $R$ sum over: step 3’s $W$ is summed off the product grid, step 4’s $W_{\tau}$ off the τ-tilted grid (the subscript names the grid the sum runs over — it is not a power). For example the 1-0 cell (a τ cell) goes $J[1][0]: 0.1095 \to 0.1009 \quad (\tau(1,0) = 1 + \lambda_a \rho = 0.8957)$ while the re-split of λₕ vs λₐ moves the untouched cells by exactly enough that the aggregate lands back on the targets. The side-by-side grids at the end of this step show that recomposition cell by cell.

Unroll the search — every bisection iteration

Outer loop — size. Each row takes the bracket midpoint as a trial $\lambda_{\text{total}}$, runs the full inner bisection (second table) to find the λₕ that restores the cover target on that trial’s τ-grid, then compares the resulting fair $P(\text{over})$ with $p^{\ast}_{\text{over}} = 0.5000$ to decide which half of the bracket survives:

k	lo	hi	trial λ_total	inner λₕ	fair P(over)	move
1	0.020000	8.000000	4.010000	2.314064	0.75390658	↓
2	0.020000	4.010000	2.015000	1.314016	0.32573724	↑
3	2.015000	4.010000	3.012500	1.807982	0.57445562	↓
4	2.015000	3.012500	2.513750	1.559643	0.45654525	↑
5	2.513750	3.012500	2.763125	1.683477	0.51759954	↓
6	2.513750	2.763125	2.638437	1.621478	0.48755270	↑
7	2.638437	2.763125	2.700781	1.652457	0.50270268	↓
8	2.638437	2.700781	2.669609	1.636962	0.49515858	↑
9	2.669609	2.700781	2.685195	1.644708	0.49893845	↑
10	2.685195	2.700781	2.692988	1.648582	0.50082253	↓
11	2.685195	2.692988	2.689092	1.646645	0.49988098	↑
12	2.689092	2.692988	2.691040	1.647614	0.50035188	↓
13	2.689092	2.691040	2.690066	1.647130	0.50011646	↓
14	2.689092	2.690066	2.689579	1.646888	0.49999873	↑
15	2.689579	2.690066	2.689822	1.647009	0.50005760	↓
16	2.689579	2.689822	2.689701	1.646948	0.50002816	↓
17	2.689579	2.689701	2.689640	1.646918	0.50001345	↓
18	2.689579	2.689640	2.689609	1.646902	0.50000609	↓
19	2.689579	2.689609	2.689594	1.646895	0.50000241	↓
20	2.689579	2.689594	2.689586	1.646891	0.50000057	↓
21	2.689579	2.689586	2.689583	1.646889	0.49999965	↑
22	2.689583	2.689586	2.689585	1.646890	0.50000011	↓
23	2.689583	2.689585	2.689584	1.646890	0.49999988	↑
24	2.689584	2.689585	2.689584	1.646890	0.49999999	✓

Inner loop — split. At the converged $\lambda_{\text{total}} = 2.689584$, each row prices the τ-grid at $(\lambda_h,\; \lambda_{\text{total}} - \lambda_h)$ and checks fair $P(\text{cover})$ against $p^{\ast}_{\text{cover}} = 0.5000$:

j	lo	hi	trial λₕ	fair P(cover)	move
1	0.010000	2.679584	1.344792	0.35759231	↑
2	1.344792	2.679584	2.012188	0.67290199	↓
3	1.344792	2.012188	1.678490	0.51520853	↓
4	1.344792	1.678490	1.511641	0.43527488	↑
5	1.511641	1.678490	1.595066	0.47509906	↑
6	1.595066	1.678490	1.636778	0.49513575	↑
7	1.636778	1.678490	1.657634	0.50516984	↓
8	1.636778	1.657634	1.647206	0.50015194	↓
9	1.636778	1.647206	1.641992	0.49764360	↑
10	1.641992	1.647206	1.644599	0.49889772	↑
11	1.644599	1.647206	1.645902	0.49952482	↑
12	1.645902	1.647206	1.646554	0.49983838	↑
13	1.646554	1.647206	1.646880	0.49999516	↑
14	1.646880	1.647206	1.647043	0.50007355	↓
15	1.646880	1.647043	1.646961	0.50003436	↓
16	1.646880	1.646961	1.646921	0.50001476	↓
17	1.646880	1.646921	1.646900	0.50000496	↓
18	1.646880	1.646900	1.646890	0.50000006	✓

move: ↑ = fair below target → keep the upper half (lo = mid) · ↓ = above → keep the lower half (hi = mid) · ✓ = |fair − target| < 1e-7 (the dc-stage tolerance; 60-iteration cap). The two final rows land on (1.6469, 1.0427) — step 4’s pair in the ⟹ line above. Every earlier outer row ran this same inner loop against its own trial λ_total; the inner table shown is the one belonging to the winning size.

Reading the repair: each miss has its own knob — a Total miss moves the size (step 2’s bisection on λ_total), a Handicap miss moves the split (step 3’s on λₕ). Here the tilt knocked the cover target off harder (by 0.0080, vs 0.0038 for the over) — ρ < 0 parks its extra mass on the draw cells 0-0/1-1, and on a home -0.50 line a draw is a lost cover — so the correction lands mainly in the split: λₕ +0.0090, λₐ −0.0243, λ_total −0.0152.

Both stages so far are the same nested solve (steps 2–3), each run on its own grid:

Stage	λₕ	λₐ	λₕ+λₐ	Δ cover	Δ over
Poisson	1.6378	1.0670	2.7048	1.9e-9	1.4e-9
Dixon-Coles	1.6469	1.0427	2.6896	6.0e-8	8.0e-9

These rows are the score-grid panel’s λ chips; the Δ’s are each stage’s achieved |model − target| residuals — the convergence check. One optional constraint remains before the markets price — step 5.

Side by side, the tilt becomes visible — the low-score corner of both grids, every cell shaded by how Dixon-Coles moved it (green = up, red = down; hover a cell for the exact before → after, and click any Dixon-Coles cell for its full derivation from the Poisson pieces):

Poisson (λₕ 1.6378 · λₐ 1.0670)

	A=0	A=1	A=2	A=3	A=4	A=5
H=0	0.0669	0.0714	0.0381	0.0135	0.0036	0.0008
H=1	0.1095	0.1169	0.0624	0.0222	0.0059	0.0013
H=2	0.0897	0.0957	0.0511	0.0182	0.0048	0.0010
H=3	0.0490	0.0523	0.0279	0.0099	0.0026	0.0006
H=4	0.0201	0.0214	0.0114	0.0041	0.0011	0.0002
H=5	0.0066	0.0070	0.0037	0.0013	0.0004	0.0001

Dixon-Coles, re-solved (λₕ 1.6469 · λₐ 1.0427)

	A=0	A=1	A=2	A=3	A=4	A=5
H=0	0.0802	0.0596	0.0372	0.0129	0.0034	0.0007
H=1	0.1009	0.1293	0.0613	0.0213	0.0056	0.0012
H=2	0.0928	0.0968	0.0504	0.0175	0.0046	0.0010
H=3	0.0509	0.0531	0.0277	0.0096	0.0025	0.0005
H=4	0.0210	0.0219	0.0114	0.0040	0.0010	0.0002
H=5	0.0069	0.0072	0.0038	0.0013	0.0003	0.0001

mass up vs Poisson mass down the four τ cellsshade ∝ √|Δ| · both grids share the same shading · shown 0–5, higher cells barely move

5 · Lift the diagonal to the 1X2 draw (φ · Method A) — optional

The 1X2 is a three-way book, so Shin stripped it in step 1: with implied $q_i = 1/d_i$ and overround $Q = \textstyle\sum q_i$, Shin’s insider fraction $z$ deflates each side to

$$p_i(z) \;=\; \frac{\sqrt{z^{2} + 4(1-z)\,q_i^{2}/Q\,} - z}{2(1-z)} \qquad \text{bisect } z \in [0,1) \text{ until } \textstyle\sum p_i = 1 \;\Rightarrow\; z = 0.0308$$

giving fair $P(\text{home})\, 0.5060$, $\text{draw}\, 0.2995$, $P(\text{away})\, 0.1945$. The handicap already fixes the win sides; the draw — the grid’s diagonal $\textstyle\sum_k J[k][k]$ — is the new constraint. Multiplying every equal-score cell by $(1 + \varphi)$, renormalizing the grid to $\textstyle\sum J = 1$, and re-solving $(\lambda_h, \lambda_a)$ so the Handicap and Total still hold, a third bisection lands φ on the draw (Method A: one flat factor across the whole diagonal, i.e. per-cell weights $w_k = 1$). The “two-input draw” below is the φ = 0 grid’s diagonal mass — the Dixon-Coles stage’s draw chip in the score-grid panel:

$$\underbrace{0.2706}_{\text{two-input draw}} \;\xrightarrow{\;\varphi \,=\, 0.2042\;}\; \underbrace{0.2995}_{\text{1X2 draw target}} \qquad \Delta = 2.0 \times 10^{-10}$$

Unroll the φ search — every bisection iteration

Third loop — the draw. Each row lifts every equal-score cell by $(1 + \varphi)$ at its trial φ, renormalizes, re-solves $(\lambda_h,\, \lambda_a)$ against the Handicap and Total targets — the entire two-level bisection unrolled in step 4, now on the lifted grid — and reads the solved grid’s diagonal $\textstyle\sum_k J[k][k]$ against the draw target $0.2995$. Bracket: the two-input draw 0.2706 < target → inflate it: φ ∈ [0.0000, 1.0000]:

i	lo	hi	trial φ	λₕ	λₐ	draw	move
1	0.000000	1.000000	0.50000000	1.851442	0.980329	0.333069327	↓
2	0.000000	0.500000	0.25000000	1.752676	1.011626	0.305280237	↓
3	0.000000	0.250000	0.12500000	1.700712	1.027253	0.288930801	↑
4	0.125000	0.250000	0.18750000	1.726920	1.019454	0.297333627	↑
5	0.187500	0.250000	0.21875000	1.739853	1.015542	0.301362045	↓
6	0.187500	0.218750	0.20312500	1.733400	1.017499	0.299361869	↑
7	0.203125	0.218750	0.21093750	1.736630	1.016521	0.300365439	↓
8	0.203125	0.210938	0.20703125	1.735016	1.017010	0.299864494	↓
9	0.203125	0.207031	0.20507813	1.734208	1.017255	0.299613402	↓
10	0.203125	0.205078	0.20410156	1.733804	1.017377	0.299487680	↑
11	0.204102	0.205078	0.20458984	1.734007	1.017316	0.299550541	↓
12	0.204102	0.204590	0.20434570	1.733906	1.017346	0.299519081	↓
13	0.204102	0.204346	0.20422363	1.733855	1.017361	0.299503397	↑
14	0.204224	0.204346	0.20428467	1.733880	1.017354	0.299511273	↓
15	0.204224	0.204285	0.20425415	1.733868	1.017358	0.299507338	↓
16	0.204224	0.204254	0.20423889	1.733861	1.017359	0.299505365	↓
17	0.204224	0.204239	0.20423126	1.733858	1.017360	0.299504384	↓
18	0.204224	0.204231	0.20422745	1.733857	1.017361	0.299503893	↓
19	0.204224	0.204227	0.20422554	1.733856	1.017361	0.299503661	↓
20	0.204224	0.204226	0.20422459	1.733856	1.017361	0.299503495	↑
21	0.204225	0.204226	0.20422506	1.733856	1.017361	0.299503578	↑
22	0.204225	0.204226	0.20422530	1.733856	1.017361	0.299503619	↑
23	0.204225	0.204226	0.20422542	1.733856	1.017361	0.299503640	↓
24	0.204225	0.204225	0.20422536	1.733856	1.017361	0.299503630	↑
25	0.204225	0.204225	0.20422539	1.733856	1.017361	0.299503635	↓
26	0.204225	0.204225	0.20422538	1.733856	1.017361	0.299503632	↓
27	0.204225	0.204225	0.20422537	1.733856	1.017361	0.299503631	✓

move: ↑ = draw below target → keep the upper half (lo = mid) · ↓ = above → keep the lower half (hi = mid) · ✓ = |draw − target| < 1e-9 (48-iteration cap). The ✓ row is the stage’s result: φ = 0.2042 with the re-solved pair (1.7339, 1.0174) — the 1X2 φ chips in the grid panel and the last row of the table below. Every row re-ran the whole step-4 nest at its own trial φ.

The φ stage re-runs the same nested solve with this third constraint attached, completing step 4’s per-stage table:

Stage	λₕ	λₐ	λₕ+λₐ	φ	Δ cover	Δ over	Δ draw
Poisson	1.6378	1.0670	2.7048	—	1.9e-9	1.4e-9	—
Dixon-Coles	1.6469	1.0427	2.6896	—	6.0e-8	8.0e-9	—
1X2 φ	1.7339	1.0174	2.7512	0.2042	1.7e-9	2.1e-8	2.0e-10

Every market below prices off the last row’s grid; Δ draw is the φ bisection’s achieved residual.

Consistency — the Handicap owns the tilt, so the 1X2 win legs are a cross-check, not a constraint: model P(home win) 0.5000 vs 1X2 0.5060, P(away win) 0.2005 vs 0.1945. A large gap means the three quotes disagree (different books or times). One draw number fixes the diagonal’s level only — the 2-2/3-3/4-4 shape rides on the flat wₖ; pin it with correct-score quotes. Flat wₖ also lifts only even totals (every k-k totals 2k), so the fit nudges Odd/Even and Exact-Total toward even — Method A is not draw-isolated. ρ and φ both lift 0-0/1-1, so φ absorbs part of ρ — the two are not independently interpretable; flip the Dixon-Coles and 1X2 φ stages in the grid panel to see φ’s residual move.

6 · Read “Correct Score” off the score grid

With λₕ, λₐ and φ fixed, the last stage’s grid $J[h][a]$ (step 5) is fully determined — the score-grid panel above — and every market probability is just a sum of its matching cells. Each outcome is one grid cell J[h][a]; scores beyond the board cap (4-4 full match, 3-3 halves) fold into “AOS”.

Outcome	Σ from grid	Fair	Priced dec
0-0	0.0867	0.0867	11.139
0-1	0.0515	0.0515	18.615
0-2	0.0317	0.0317	29.890
0-3	0.0107	0.0107	83.286
0-4	0.0027	0.0027	269.624
1-0	0.0953	0.0953	10.142
1-1	0.1430	0.1430	6.784
1-2	0.0549	0.0549	17.468
… 18 more

Raw masses are normalized to Σ = 1, then inflated by Shin to overround 1 + 0.05 — the insider model run forward: $q_i = S \cdot t_i$ with $t_i = \sqrt{p_i((1-z)p_i + z)}$ and $S = \textstyle\sum t_i$, bisecting $z$ until $\textstyle\sum q_i = 1.05$ (this run: $z = 0.0021$).

7 · Implementation notes — nesting, brackets, normalization

targets   tH = powerStrip(HCP pair)      # fair P(home covers L)         — step 1
          tT = powerStrip(Total pair)    # fair P(over T)                — step 1
          tD = shinStrip(1X2).draw       # fair P(draw); Method A only   — step 5

bisect φ ∈ (−1, ∞)                       # outermost; skip (φ = 0) with no 1X2
  bisect λ_total ∈ [0.02, 8]             # size — step 2
    bisect λₕ ∈ (0, λ_total)             # split — step 3; λₐ = λ_total − λₕ
      J = Poisson(λₕ, λₐ) on 0…N         # ×τ + renorm (DC — step 4); φ stage: diag ×(1+φ) + renorm
    until fairHcp(J, L) = tH             # line rules — step 2
  until fairTot(J, T) = tT
until Σₖ J[k][k] = tD

The grid panel’s stages run this same nest with the adjustments off/on — Poisson: neither, Dixon-Coles: τ only, 1X2 φ: both. Every residual is monotone in its knob — cover rises with λₕ, over with λ_total, draw with φ — so plain bisection converges: 48–60 iterations per loop at tolerances 1e-9…1e-7 (the round-trip Δp above is this run’s achieved error). The φ bracket doubles upward from [0, 1], or flips to (−1, 0] when the two-input draw already exceeds the 1X2 target. The grid truncates at N = 5: the plain Poisson grid is not renormalized, the τ- and φ-adjusted grids are, and every market group renormalizes its outcome masses to Σ = 1 before its margin is applied — that absorbs the truncation loss. The executable reference for all of the above is docs/src/lib/odds/core.ts + markets.ts, held to the original simulator by the golden-master test in docs/src/lib/odds/__tests__/.